If $K:=\mathbb Q\left(\sqrt{-3}\right)$ and $R$ is the ring of integers of $K$, then $R^{\times}=\mathbb Z\big/6\mathbb Z$ 
How to show that if $K:=\mathbb Q\left(\sqrt{-3}\right)$ and $R$ is the ring of integers of $K$, then the group of units $R^{\times}=\mathbb Z\big/6\mathbb Z$

Now since $-3\equiv1\mod 4$ the ring of integers are $\mathbb Z\left[\frac{1+\sqrt{-3}}{2}\right]$
So any element in the ring is of the form $a+b\left(\frac{1+\sqrt{-3}}{2}\right)$ with $a,b\in\mathbb Z$
but I can always find another $2$ elements $\tilde{a},\tilde{b}\in\mathbb Z$ with the same parity such that 
$\displaystyle a+b\left(\frac{1+\sqrt{-3}}{2}\right)=\frac{\tilde{a}+\tilde{b}\sqrt{-3}}{2}$
Now the norm is easier to examine, if I set it equal to $1$;
$N(\frac{\tilde{a}+\tilde{b}\sqrt{-3}}{2})=\frac{\tilde{a}^2+3\tilde{b}^2}{4}=1$
$\implies\tilde{a}=\pm2,\tilde{b}=0\quad$ or $\quad\tilde{a}=\pm1,\tilde{b}=\pm1$
So there are $6$ possibilities, but how is it isomorphic to $\mathbb Z\big/6\mathbb Z$ ?
 A: Clearly $R^{\times}$ is an abelian group and you just found out that it has order $6$. But the only abelian group of order $6$ is the cyclic group on $6$ elements...

Indeed, the fundamental theorem of finitely generated abelian groups and $\#R^{\times} = 6$ imply that $R$ is a direct sum of primary cyclic groups. Since $6 = 2 \cdot 3$, the only possibility is
$$
R^{\times} \simeq \Bbb{Z}/2\Bbb{Z} \oplus \Bbb{Z}/3\Bbb{Z} \simeq \Bbb{Z}/6\Bbb{Z}
$$
where the second isomorphism is due to the Chinese remainder theorem.

Element-wise, observe that
$$
\left(\frac{1 + \sqrt{-3}}{2}\right)^2 = \frac{-2+2\sqrt{-3}}{4} = \frac{-1+\sqrt{-3}}{2}
$$
and that
$$
\frac{1 + \sqrt{-3}}{2} \, \frac{-1+\sqrt{-3}}{2} = \frac{-4}{4} = -1
$$
Since $-1$ has order $2$, it follows that $\frac{1 + \sqrt{-3}}{2}$ has order $6$ (and that $\frac{-1+\sqrt{-3}}{2}$ has order $3$).
A: I provide an alternative solution with a bit less appeal to abstract results and a focus on a concrete and strictly number-theoretic approach. Note the norm is most easily written as

$$N(a+b\zeta_3)=(a+b\zeta_3)(a+b\zeta_3^2)=a^2+b^2-ab$$

since $\zeta_3^2+\zeta_3+1=0$. The Cauchy-Schwarz inequality shows that this is always positive, so we need only examine when $a^2+b^2-ab=1$, and if $a$ and $b$ have opposite signs, the norm is $\ge 3$ as then $a^2+b^2-ab\ge 1+1+1$. So we may as well assume $a,b\ge 0$ and know that we get twice as many units as satisfy this by multiplying $a+b\zeta_3$ by $-1$. Since

$$a\ge b\ge 0\implies a^2+b^2-ab=b^2+a(a-b)\ge b^2.$$

(the last inequality follows from $a-b\ge 0$)
As such $b\in\{0,1\}$ are all we need check, and so also $a\in\{0,1\}$ also, by size considerations. The total pairs being $(a,b)\in \{(0,0),(1,0),(1,1),(0,1)\}$. Clearly $(0,0)$ is out, but the others are all seen to work, producing $3$ units apiece. Since we noted earlier we could double this number by negating all units arising in this way, there are $6$ in all.
We can also find a generator quite explicitly. Since $\zeta_3$ satisfies $\zeta_3^3=1$ and no smaller power works, then we know that $(-\zeta_3)^6=1$ and further that
$$(-\zeta_3)^3=(-1)^3(\zeta_3)^3=-1$$
so that it's order is exactly $6$. Indeed, if we think about it, we can see that $-\zeta_3=\zeta_6$, a primitive $6^{th}$ root of unity, but this is not necessary. Since there are $6$ units, and $-\zeta_3$ has order $6$ and is clearly a unit, the group is generated by this element.
